AUTOMOTIVE AND HEAVY EQUIPMENT PROGRAMS
LEVEL I
ENGINEERING SCIENCE
August, 2022
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
Introduction
This module has specifically been prepared for Instrumentation Diploma level I open distance
learning program. It has been prepared with the view of making self-study in engineering science
easy in readiness for the first level exams. This is mainly in view of the fact that the program
does not provide enough time for students to go through all the topics with the trainer(s) in class.
Consistent practice normally brings improvement which later leads to success. Therefore, as a
student under this program, do your best to manage your busy schedule in a way that will enable
you to study the lessons, and successfully answer the questions following each topic. It is
expected that after studying this subject, you will be adequately equipped with the necessary
knowledge and skills so that you can proficiently apply scientific principles in solving problems in
your field. However, remember never to duplicate, in whole or in part, this document, and help
NORTEC give you the knowledge you deserve without any complications.
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Contents
Topic
Page
1.
Fundamental Quantities
3
2.
Linear motion
11
3.
Angular motion
16
4.
Projectiles
21
5.
Momentum
28
6.
Newton’s laws of motion
33
7.
Centroids
37
8.
Equilibrium
43
9.
Friction
52
10. Work, energy and power
57
11. Heat
61
12. Temperature conversions
61
13. Linear expansion
63
14. Sensible heat and latent heat
66
15. Gas laws
69
16. Illumination
72
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FUNDAMENTAL QUANTITIES
The term fundamental refers to something which may be serving as the base or foundation. In
other terms, a fundamental quantity can be considered to be a quantity of primary importance.
The three fundamental quantities which are commonly used include mass, length and time. The
S.I. units for each of them are as follows,
➢ Mass – kilogram (kg)
➢ Length – meter (m)
➢ Time – second (S)
This system of measuring units which uses meters, kilograms and seconds is known as the
‘mks’ system of units. Previously the system that was used was the centimetre, gram, second –
‘cgs’.
SI Base units
In Engineering there are seven base units known and these form the basis for other units
derived. The table below shows the quantities involved and their respective units together with
the symbols employed.
Physical quantity
symbol
SI unit
Unit symbol
Dimension
l
Meter
m
L
2. Mass
m
Kilogram
kg
M
3. Time
t
Second
s
T
4. Electric current
I
Ampere
A
A
5. Quantity of a
substance
n
Mole
mol
n
6. Temperature
T
Kelvin
K
θ
7. Luminous intensity
IL
Candela
cd
I
1. Length
Unit conversions
Using multiples which are all powers of ten with their names or prefixes, units can be converted
to more convenient figures. The table below shows the multiples and their names in descending
order. It is expected that these are mastered and at least 90% to 100% of them should be
known to the extent of applying them correctly.
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Engineering science for Automotive & HER Programs Level I
prefix
Multiplying factor
24
Symbol
yotta
10
Or (1,000,000,000,000,000,000,000)
Y
Zetta
1021 or (1,000,000,000,000,000,000,000)
Z
Exa
10 18 or (1,000,000,000,000,000,000)
E
Peta
10 15 or (1,000,000,000,000,000)
P
Tera
10 12 or (1,000,000,000,000)
T
Giga
Mega
10
9
or (1,000,000,000)
10 6
or (1,000,000)
G
M
kilo
10 3 or (1000)
k
hecto
10 2 or (100)
h
deca
10 1 or (10)
da
deci
10 −1 or
centi
10 −2
milli
10 −3
micro
nano
pico
femto
atto
1
10
 1 

or 
 100 
 1 

or 
 1000 

C
m
1


 1,000,000 
μ
1


 1,000,000,000 
n
1




 1,000,000,000,000 
p
10 −6
or 

10 −9 or 
10 −12 or
d
10 −15 or 
1


1
,
000
,
000
,
000
,
000
,
000


f
1


 1,000,000,000,000,000,000 
a
1
)
1,000,000,000,000,000,000,000
z

10 −18 or 
zepto
10-18 or (
yocto
10-24 or (1,000,000,000,000,000,000,000)
1
Prepared by Mr. Mwaba Ernest
y
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Engineering science for Automotive & HER Programs Level I
EXAMPLES
1. Convert each of the following to the stated units;
(a) 1.5GN to N
(b) 0.05Hz to µHz
(c) 3500mA to A
(d) 20kW to W
(e) 15.8MV to V
(f)
2.5cm to mm
Solutions
(a) 1GN ⟶ 109 N
1.5GN⟶ 𝒳
Cross multiplying and making 𝓍 the subject of the formula,
1.5 × 109
𝔁=
= 1.5 × 𝟏𝟎𝟗 𝐍 or 1,500,000,000N
1
∴ 1.5GB ≡ 1.5 × 𝟏𝟎𝟗 𝐍
(b) 1µHz ⟶ 10−6 Hz
𝔁 ⟶ 0.05 Hz
10−6 𝓍 = 0.05
0.05
𝓍 = 10−6 = 50,000µHz
∴ 0.05Hz ≡50,000µHz
(c) 1mA ⟶ 10−3 A
3500mA ⟶ 𝓍
1 × 𝔁 = 3500 × 10−3
∴ 3500mA ≡ 3.5A
(d) 1kW ⟶ 103 W
20kW ⟶ 𝓍
Again cross multiplying and rearranging gives,
1× 𝓍 =20 × 103
∴ 20kW ≡ 20,000W
2. Determine the equivalent unit quantity for each of the following;
(a) 0.05kB/s to B/s
(b) 2N/mm2 to MN/m2
(c) 360km/h to m/s
(d)
2.5cm2 to m2
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Solutions
(a) Never mind because you are ready to be mature and no need to use the primitive
arrows,
1kB ⟹ 1000B
0.05kB×1000
⟹
S
The units only serve the purpose of showing how the conversion is taking place and
they can now be omitted in our finishing work.
0.05 ×1000
=
1
∴ 0.05kB/s ≡ 50B/s
(b) 2N/mm2 to MN/𝑚2
Let’s get straight to work,
2N
mm2
=
2N
106
×
(1000)2
mm2
After seeing how the conversion is taking place, the units can be omitted as we deal
with the numbers
⟹
2×1000,0000
106
= 2MN/m2
∴ 2N/𝐦𝐦𝟐 ≡ 2MN/𝐦𝟐
(c) 360km/h to m/s
⟹
360km ×1000
1
⟹
360×1000
3600
×
1
60×60h
= 100m/s
∴ 360km/h ≡ 100m/s
(d) 1m2 ⟹ 1002 cm2
?
⟹ 2.5cm2
⟹
2.5
10,000
= 2.5× 𝟏𝟎−𝟒
∴ 2.5𝐜𝐦𝟐 = 2.5× 𝟏𝟎−𝟒 𝐦𝟐
EXERCISE
1. Convert each of the following;
(a) 100kV to V
(Ans100,000V)
(b) 0.02A to mA
(Ans: 20mA)
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(c)
(d)
2. Carry
(a)
(b)
(c)
(d)
40MHz to kHz
(Ans: 40,000kHz)
90,000kB to GB (Ans: 0.09GB)
out the following conversions;
2KB/s to MB/s
(Ans: 0.002MB/s)
300kB/min to B/s
(Ans: 5,000B/s)
72km/h to m/s
(Ans: 20m/s)
2
2
5.3MN/m to N/mm
(Ans: 5.3N/mm2)
3. Express the following to the stated units,
(a) 50,000mm3 to m3 (Ans: 5× 10−5 m3 )
(b) 20.5m2 to mm2
(Ans: 20.5× 106 )
(c) 40cm2 to mm2
(Ans: 4000mm2)
3
3
(d) 2g/cm to kg/m (Ans: 2000kg/m3)
4. Calculate the equivalent quantity of the specified unit for each of the following;
(a) 200 000µF to F (Ans: 0.2F)
(b) 50mm to km (Ans: 5× 10−5km)
(c) 47.8liters to m3 (Ans: 0.0478m3)
Dimensional analysis
This is a kind of algebra which deals with the dimensions of units and the aim of this
algebra is to help in verifying the dimensional correctness of a stated formula or simply to
check equations.
An equation is dimensionally correct if all the dimensions on the right hand side of an equal
sign also appear on the left hand side of an equal sign with the same powers. When a
constant appears on any or both sides of an equal sign while the rest is balanced, then an
equation may still be considered to be dimensionally correct because constants do not affect
the dimensions of an expression.
Examples
1. Verify the dimensional correctness of the following formulas;
(a)
V = u + at
(b)
V=
(c)
time =
2 gh
displacment
acceleration
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Solutions
m m m

= +  2  s
s
s s

1. (a)
L L  L

= +
T 
T T  T 2

L L L
= +
T T T
L
L
=2
T
T
The number 2 is a constant like any other number in this analysis, ∴ the equation is
dimensionally correct
1
m  m
2
= 2 2  m
s  s

(b)
1
L  L
2
= 2 2  L 
T  T

1
L  L2  2
= 2  According to the laws of indices this can be simplified as follows,
T  T2
1
 
 
L
L2
= 22
T
T2
or
L
=
T
( 2)
1
2
L2
T

1
2
2

L L
=
2
T T
L L
=
2
T T
The term √𝟐 is a constant and therefore, the equation is dimensionally correct
(c) T =
L
T=
T2
L
T2
T =
L
T2
L
T=T
The equation is dimensionally correct
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2. State whether the following formulas are dimensionally correct or not.
mass time
velocity
(a)
Force =
(b)
Acceleration =
velocity displacement
time
Solutions;
m
m
 kg = kg  s 
2
s
s
L
L
 M = M T 
T
T
2. (a)
L.M
T
= M .T 
T
L
L.M M .T 2

T
L
∴The equation is dimensionally incorrect.
(b)
m m
= ms
s
s2
L
L
1
=  L
2
T
T
T
2
L
L
 2
2
T
T
∴The equation is dimensionally incorrect
EXERCISE
1. Prove the dimensional correctness of the following formulas
a. V 2 = U 2 + 2aS
U +V 
t
 2 
b.
S= 
c.
Mass =
density area
displacement
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d.
Velocity =
e.
Force =
acceleration  time
displacement
power
velocity
2. State whether each of the following equation is dimensionally correct or not;
velocity
acceleration
a.
Time =
b.
mass  acceleration = pressure  area
c.
Force  volume
= power  time
area
3. Mass × velocity = acceleration × time
4. Pressure × area = mass × time
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LINEAR MOTION
The word linear comes from the term line; therefore linear motion occurs when a body is
moving in a straight line. In the study of linear motion the following terminology may be of
importance,
Distance;
This is defined as the length between two points. It is also tentatively defined as how far it is
between any two given points. In the measurement of distance there is no consideration on the
aspect of direction, and thus distance is a scalar quantity.
Displacement:
This is the distance covered in a specified direction. It is measured in meters (m) and denoted
by the letter S.
Speed;
This is the rate of change of distance or the change of distance with respect to time. In other
terms it does not require the direction to be specified and it is also a scalar quantity.
speed =
dis tan ce
time
Velocity;
Velocity is defined as the rate of change of displacement. It can also be defined as the change
of displacement with respect to time. Mathematically, it is presented as follows,
Velocity =
displacement
time
However, this can be written using symbols as follows,
V =
S
(m / s)
t
Where, S = displacement in meters
t = time in seconds
V = velocity in m/s
Acceleration;
This is the rate of change of velocity. It is also defined as the change in velocity with respect to
time. Thus,
a=
V
( m/s2 ), but ΔV = V – u
t
∴a=
v−u
t
Where, v = the final velocity, u = Initial velocity
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Equations of motion
There are four main equations of linear motion of a uniformly accelerated body. The symbols
used in these equations have already been discussed above.
1. V = u + at
2. V 2 = u2 + 2aS
1
3. S = ut + 2 at 2
𝑈+ 𝑉
)t
2
4. S = (
EXAMPLES
1. A body moving at 12m/s is accelerated uniformly until the final velocity becomes
32m/s. If the time taken for the acceleration is 20seconds, calculate;
(a) The uniform acceleration
(b) The displacement covered during the acceleration period.
Data
u = 12m/s; v = 32m/s; t = 20sec,
Solution
(a) a =
v−u
t
=
32−12
20
a = 1m/s2
u+v
)t
2
(b) s = (
12+32
) 20
2
S= (
∴ 𝐒 = 𝟒𝟒𝟎𝐦
2. A particle initially running at 1× 1014m/s is decelerated uniformly until the speed
becomes 2 × 1012m/s in 200 μS. Determine;
(a)
The uniform deceleration
(b)
The displacement covered
(c)
The time taken to bring the particle to rest from the final speed using the
deceleration in (a)
DATA
U =1 × 1014 m/s; v = 2 × 1012m/s; t = 200 × 10−6 = 2 × 10−4s
Solutions
(a)
a=
𝐯−𝐮
𝐭
=
(2×1012 )− (1×1014 )
2×10−4
a = −𝟒. 𝟗 × 𝟏𝟎𝟏𝟕 𝒎/𝒔𝟐
(b)
V+u
)t
2
S=(
=[
(1×1014 ) + (2×1012 )
2
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] (2 × 10−4 )
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Engineering science for Automotive & HER Programs Level I
S =𝟏. 𝟎𝟐 × 𝟏𝟎𝟏𝟎 𝐦
(c)
t=
v−u
a
0 − (2×1012 )
=
−4.9×1017
= 𝟒. 𝟎𝟖 × 𝟏𝟎−𝟔 𝐒
3. A body starts from rest and attains a speed of 40m/s in 20seconds. It maintains this
speed for 30seconds and it is then brought to rest at a uniform rate in 540m. For the
motion of this body;
(a)
Draw the velocity – time graph
(b)
Determine the uniform acceleration
(c)
Determine the uniform retardation
(d)
Calculate the total displacement for the whole journey
Solutions
(a)
The velocity time graph is as illustrated below
V
40m/s
S1
S2
t1=20s
v−u
t
40− 0
20
t2= 30s
(b)
a=
(c)
v = 0; u = 40m/s; s = 540m; a=?
=
S3 = 540m
Time(s)
t3
= 2m/𝐬 𝟐
v 2 = u2 + 2as
a=
v2 −u2
2s
=
02 − 402
2×540
a = - 1.48m/𝒔𝟐
(d)
u+v
)t
2
s1 = (
0+40
)
2
=(
× 20 = 400m and s2 = vt = 40 × 30 = 1200m
s3 = 540m
∴ Total displacement, s = s1 + s2 + s3 =400 + 540 + 1200
s = 2140m
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4. A ball is thrown vertically upwards at a velocity of 80m/s from the top of a 90m high
building. If the effect of the air resistance is ignored, calculate the;
(a)
maximum height reached by the ball above the building
(b)
time taken to reach the maximum height
(c)
velocity of the ball just before hitting the ground
(d)
Duration of the whole motion
DATA
u = 80m/s; v =0; s =?; a =-g =-9.81m/s2
Solutions
(a)
v 2 = u2 + 2as
s=
v2 −u2
2a
=
02 − 802
2×(−9.81)
s = 326.198 m
(b)
t=
v−u
a
=
0−80
−9.81
∴ t = 8.15 s
(c)
u = 0; v =? a = 9.81m/s2; s = 90+326.198=416.198m
v 2 = u2 + 2as
v 2 = 02 + (2 × 9.81 × 416.198)
v = √8165.805
v = 90.36m/s
(d)
1
s2 = ut + 2 at 22
1
416.198= (0 × t) + 2 t 22
t 2 = √(
416.198
)
4.905
= 9.212sec
𝑡1 = 8.15𝑠𝑒𝑐
∴ Total time, t = t1 + t 2 = 8.15 + 9.212 = 17.36sec
EXERCISE
1. A body starts from rest and attains a speed of 72km/h in 500m. Calculate;
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(a)
(b)
(c)
Average velocity (Ans: 10m/s)
Time taken
(Ans: 50sec)
Acceleration
(Ans: 0.4m/s2)
2. An electron initially spinning round its path at 1× 109 m /s gains its energy level by
attaining a speed of 2.5 × 1011m /s in 10ms. Determine;
(a)
acceleration of the electron
(Ans: 2.49× 1013m/s2)
(b)
Displacement covered by the electron (Ans: 1.255× 109m)
3. A parcel is dropped from a height of 200m above the ground. If the air resistance is
ignored in analyzing its motion, calculate the;
(a)
Time taken to reach the ground
(Ans: 6.38sec)
(b)
Velocity with which it hits the ground (Ans: 62.64m/s)
4. A rocket starts from rest and accelerates uniformly at 25m/s2 in 5seconds. Determine;
(a)
Final velocity
(Ans: 125m/s)
(b)
Displacement covered (Ans: 312.5m)
(c)
Time taken for the acceleration due to gravity to bring it to rest from the
maximum speed.
(Ans: 12.742s)
5. A body is accelerated uniformly from 20m/s until it attains a speed of 60m/s in 100m. It
maintains this speed for a period of 45seconds and is finally brought to rest in another
62seconds after some uniform retardation. Draw the velocity – time graph and use it
where applicable to calculate the;
(a)
Uniform acceleration
(Ans: 16m/s2)
(b)
Duration of the acceleration period (Ans: 2.5sec)
(c)
Uniform retardation
(Ans: 0.968m/s2)
(d)
Duration of the whole journey
(Ans: 109.5sec)
6. A parcel of mass 2.3kg is thrown vertically upwards at 180km/h from the top of a 75m
high surface. Ignoring the effect of the air resistance, calculate the;
(a)
Time taken for the parcel to reach the maximum height
(Ans: 5.1sec)
(b)
Velocity with which it passes its starting point from maximum height (Ans:50m/s)
(c)
Maximum height reached by the parcel above the starting point (Ans: 127.42m)
(d)
Striking velocity (Ans: 63.1m/s)
7. A space craft is launched at a speed of 30,000km/h until it attains the targeted speed of
40,000km/h. If the time taken to reach the targeted speed is 3.5minutes, calculate the;
(a)
Distance covered during this period in kilometers (Ans: 2041.67km)
(b)
Acceleration of the space craft
(Ans: 13.23m/s2)
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ANGULAR MOTION
Angular motion is the movement of a body about an axis, in a circular path. The displacement
of a body moving in angular motion is measured in radians. In analyzing angular motion the
following definitions are important;
Radian;
The radian is an angle subtended at the center by an arc whose length is equal to the radius of
the circle. Diagrammatically, this can be presented as shown below,
S=r
r
1rad
1revolution ≡2π radians, and since there are 360° in one revolution, then
360°≡ 2π rads, dividing by 2π on both sides gives,
𝟏𝟖𝟎°
1rad = 𝝅 degrees or 1radian = 57.29°
Relating Revs/min ( rpm) to rad/s,
Following the above relationship between degrees and radians the rotational speed can be
converted from revs /min to rad/s using the relationship below,
𝟐𝝅𝑵
𝝎=
rad/s,
𝟔𝟎
Where, ω = angular speed in rad/s; N = angular speed in rpm or rev/min.
Angular velocity;
The angular velocity is the rate of change of angular displacement. In other terms it is the
change in angular displacement with respect to time. Thus,
𝝎 =
𝚫𝜽
𝒕
(rad/sec)
𝜽
or simply, 𝝎 = 𝒕
Where, ∆θ= change in displacement (in radians)
.
t = time in seconds
Angular acceleration, 𝜶;
This is defined as the rate of change of angular velocity. It can also be defined as the change in
angular velocity with respect to time. Thus;
𝝎𝟐 − 𝝎𝟏
(𝒓𝒂𝒅/𝒔𝟐 )
𝒕
Where, ω1=initial angular velocity in rad/s; ω2= final angular velocity in rad/s; t = time in
seconds
𝜶=
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EQUATIONS OF MOTION
The equations of angular motion are derived from the equations of linear motion. It therefore
means that each equation of angular motion has its exact equivalent in linear motion as shown
below,
LINEAR MOTION
ANGULAR MOTION
1. v = u + at
1.
2. v 2 = u2 + 2as
ω2 = ω1 + αt
2. 𝜔22 = 𝜔12 + 2𝛼𝜃
1
3. s = ut + 2 at 2
3. θ = ω1 t +
U+V
)t
2
4. S = (
1
αt 2
2
ω1 + ω2
)t
2
4. θ = (
Linear and angular motion
Angular displacement, angular velocity and angular velocity can be converted to their linear
motion equivalent using the following equations,
Angular displacement to linear displacement;
(i)
S = rθ (m),
where, r = radius in meters, S = linear displacement
Angular velocity to linear velocity;
(ii)
V = ω2r (m/s)
And U= ω1r,
Where, V = final linear velocity; U=initial linear velocity
(iii)
Angular acceleration to linear acceleration;
a = 𝜶r (m/s2)
EXAMPLES
1. A disc initially rotating at 200rev/min is accelerated uniformly until the final speed
becomes 980rpm. If the acceleration period lasts for 50seconds, calculate;
a. The angular acceleration
b. The angular displacement covered during the 50seconds
c. The number of revolutions turned through during the 50seconds
DATA
ω1=
2πN1
60
=
2π×200
60
= 20.944rad/s; ω2 =
Prepared by Mr. Mwaba Ernest
2πN2
60
=
2π × 980
60
= 102.625rad/s; t = 50sec
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Engineering science for Automotive & HER Programs Level I
Solution
a. ω2 = ω1 + αt
⇒𝛼=
𝜔2 − 𝜔1
𝑡
ω2 + ω1
)t
2
b. θ = (
=
102.625 −20.944
50
= (
= 1.634m/s2
102.625 + 20.944
)×
2
50 = 𝟑𝟎𝟖𝟗. 𝟐𝟐𝟓𝒓𝒂𝒅𝒔
c. The number of revolutions,
n=
θ
2π
=
𝟑𝟎𝟖𝟗.𝟐𝟐𝟓
2π
= 491.665revs
2. A wheel covers 300revolutions in 10seconds until its speed becomes 1300rev/min. If the
wheel has a diameter of 150mm, calculate;
a. The initial angular velocity of the wheel
b. The angular acceleration experienced by the wheel
c. The linear acceleration
d. The initial peripheral speed
DATA
θ
n = 300revs; n =
2π
⇒ θ = 2π n = 2π × 300 = 1884.956rads
t = 10sec
2π N
2π×1300
ω2 = 60 2 = 60 = 136.136rad/s
d = 0.15m ⇒ r =
0.15
= 0.075m
2
Solutions
𝝎𝟐 + 𝝎𝟏
)𝒕
𝟐
(a) 𝜽 = (
Making ω1 the subject of the formula, gives;
2θ
ω1 =
− ω2
t
=
2 × 1884.956
− 136.136
10
∴ 𝝎𝟏 = 𝟐𝟒𝟎. 𝟖𝟓𝟓𝒓𝒂𝒅/𝒔
(b)
There is no running away from the angular velocity we found in (a) because all
equations contain the initial angular velocity.
ω2 = ω1 + αt
𝛼=
𝜔2 −𝜔1
𝑡
=
136.136−240.855
10
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Engineering science for Automotive & HER Programs Level I
𝜶 = −𝟏𝟎. 𝟒𝟕𝟐𝒓𝒂𝒅/𝒔𝟐
(c)
Following the relationship between linear and angular acceleration the formula below
can be applied
a = 𝛼r
a = 10.472×0.075
a = 0.785m/s2
(d)
The initial peripheral speed refers to the initial circumferential or linear speed, thus;
U = ω1 r
U = 240.855 ×0.075
U = 18.064m/s
3.
A compact disc rotating at 1450rev/min suddenly experiences a power failure which
allows it to come to rest in 55seconds. Calculate the total linear displacement covered
during the deceleration period given that the diameter of the disc is 100mm.
DATA
2𝜋 ×1450
0.1𝑚
ω1 =
= 151.844rad/s; t=55sec; ω2= 0; r = 2 = 0.05𝑚
60
Solution
ω +ω
θ = ( 2 2 1) t
0+151.844
)×
2
=(
55
θ = 4175.71rads
S = rθ
= 0.05×4175.71
S = 208.7855m
EXERCISE
1. A wheel is accelerated from 600rpm to 2100rpm in 30seconds. If the rate at which the
speed is increased is uniform, calculate the;
(a) Angular acceleration
(Ans: 5.236rad/s2)
(b) Angular displacement covered during this time
(Ans: 4241.15rads)
(c) Number of revolutions turned through
(Ans: 675revs)
2. A rotor starts from rest and attains a speed of 3150rev/min at a rate of 40rad/s2. If the
diameter of the rotor is 800mm, calculate;
(a) Angular displacement covered by the rotor
(Ans: 1360.155rads)
(b) Number of revolutions turned through by the rotor (Ans: 216.475revs)
(c) Final peripheral speed
(Ans: 131.947m/s)
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Engineering science for Automotive & HER Programs Level I
3. The cooling fan of diameter 100mm is rotating at 900rpm when the power supply is
suddenly cut off. If the time taken for the fan to come to rest is 1.5minutes, calculate;
(a) Angular retardation
(Ans: -1.047rads/s)
(b) Angular displacement
(Ans: 4241.15rads)
(c) Linear displacement covered by a point on the circumference (Ans: 424.115m)
4. A disc starts from rest and accelerates uniformly until it attains a speed of 200rpm in
90revolutions. If the disc is then decelerated to rest from the maximum speed in
2minutes, calculate the;
(a) Angular acceleration
(Ans: 0.388rad/s 2)
(b) Time taken to reach the maximum velocity from rest
(Ans: 54s)
(c)
Uniform retardation of the disc
(Ans: -0.174rad/s2)
(d) Number of revolutions turned through during the retardation period
(Ans:200revs)
5. A drum which is initially running at 40rad/s is accelerated uniformly at a rate of 10rad/s2
until it covers a linear displacement of 420m. If the radius of the drum is 30cm, calculate
the;
(a) Final angular velocity
(Ans:240rad/s)
(b) Time taken to reach the final angular velocity (Ans: 20sec)
6. A winding cylinder is rotated at a constant speed of 150rpm. If the diameter of the
cylinder is 400mm and it is about to roll its first layer along its periphery, calculate the
length of rope wound round the cylinder in 5minutes. (Ans: 942.478m)
7. The blades of a fan which is running at 120rad/s are constructed with a diameter of
70cm. If 4200radians are covered in the deceleration of the fan uniformly for a period of
50seconds, calculate the;
(a) Final angular velocity (Ans: 48rad/s)
(b) Linear deceleration (Ans: -0.504m/s2)
8. A motor initially running at 500rpm is brought to rest after a uniform retardation of
15rad/s2. If it covers a linear distance of 50m before finally coming to rest, calculate;
(a) Time taken before the motor comes to rest (3.49sec)
(b) Diameter of the rotor for this motor
(Ans: 0.547m)
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Engineering science for Automotive & HER Programs Level I
PROJECTILES
A projectile is a body which is provided with an initial velocity and then allowed to move freely
under the action of its own weight or gravity. Some common examples of projectiles include the
following:
(i)
a ball thrown in the air
(ii)
A missile launched at some point
(iii)
A stone thrown in the air
(iv)
A motor bike jumping obstacles
A trajectory is the path followed by a projectile. A trajectory is normally expected to be
parabolic in shape and general appearance, though it is not expected that all trajectories will
have the same shape and results.
The range is the horizontal distance from the point where the projectile is launched to where
the projectile reaches the landing point or end point.
Equations of motion,
It is worth noting from the start, that a projectile can be launched at any desirable angle. This
implies that the motion of a projectile can be analysed in two portions namely, vertical motion
and horizontal motion. The same equations of linear motion can be rewritten I the two motions
as below:
Vertical motion:
a y = the acceleration due to gravity, g
V y = final velocity of body in the y-axis
(i)
V y = Usinθ – gt
(ii)
𝑉𝑦2 = U 2 sin 2 θ – 2g S y
(iii)
h= S y = Usinθt –
(iv)
h = Sy =
1
2
gt 2
1
( Usinθ + V y ) t
2
Horizontal motion:
In the X-axis the velocity is constant because the acceleration is zero, and thus
Vx = U sin
Range = S x = Ucosθt
The two equations can be derived by replacing in the four equations of linear motion the two
conditions i.e. a x = 0
Ux = Vx = U sin
Examples
1. A projectile is launched from the earth’s surface with a speed of 72km/h at an elevation
of 30º to and above the horizontal. For the motion of this projectile, calculate:
(a)
the total time of flight
(b)
the maximum height reached
(c)
the horizontal range
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Engineering science for Automotive & HER Programs Level I
20sin30
20m/s
h max
30°
20cos30
Range = 20cos30.t
Solution
(a) from the first equation, V
y
= Usinθ – gt, as U=
72km / h
= 20m /s
3 .6
at h max , V y = 0 and this is the time taken to reach maximum height, i.e
0 = Usinθ – gt
t=
U sin 
g
2U sin 
g
2  20 sin 30
t=
9.81
total time, t =
t = 2.04seconds
(b) V y
2
= U 2 sin 2 θ – 2gS y
At h max , V y = 0
0 = 20 2 sin 2 30 - 2  9.81 S y
h max =S y = 5.1m
(c) Range = Ucosθt
= (20cos30)  2.04
= 35.33m
2. A projectile is launched from the top of a 30m high building at a velocity of 30m/s and
an elevation of 45° to and above the horizontal. Calculate the;
(a)
(b)
(c)
(d)
maximum height reached from base of the building
total time of flight
Horizontal range
speed of the projectile just before reaching the ground
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Engineering science for Automotive & HER Programs Level I
data.
U y =30sin 45°; U x =30cos 45°, at h max V y = 0
30m/s
30sin45
h max
45°
30cos45
30m
V
Range
Vx = ucosθ
Solution
(a) V 2 y = U 2 sin 2 θ – 2gS y
u 2 sin 2 
Sy =
2g
i.e. After putting V y = 0 in above equation
30 2 sin 2 45
Sy=
= 22.936m
2  9.81
h max = 30 + S y ; where S y is the height above the building.
h max = 30 + 22.936
h max = 52.936m
(b) let, t 1 = time taken for projectile to move from point of projection to h max
At h max , V y = 0
V y = Usinθ – gt
0 = 30sin45 – 9.81  t 1
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Vy
24
Engineering science for Automotive & HER Programs Level I
t1 =
30 sin 45
9.81
t 1 = 2.162s
let, t 2 = the time taken for the movement from h max to the bottom, in which case,
a y = 9.81 and putting S y = 52.936m, while U y = Usinθ = 0
1 2
gt
2
1
52.936 = 0  t +  9.81 t 2
2
S y = Usinθt +
2
 52.936 

 = 3.285s
 4.905 
t2 =
Total time, t = t 1 + t 2
= 2.162 + 3.285
t = 5.55s
(c) Range = Ucosθt
Range = (30cos45)  5.55
Range = 117.733m
(d) V 2
y
= U 2 sin 2 θ – 2gS y
But U y = Usinθ = 0
Vy=
(2 gS y )
(2  9.81 52.936)
=
V y = 32.227m/s downwards
V x = Ucosθ = 30cos45 = 21.213m/s
V=
=
(V
2
y
+V 2x
(32.227
2
)
+ 21.2132
)
V = 38.582m/s
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Engineering science for Automotive & HER Programs Level I
3. A helicopter is flying horizontally at a velocity of 126km/h when it drops an object
horizontally. If the helicopter is flying at a height of 50m above the ground, calculate;
(a) The time taken for the object to reach the ground
(b) The horizontal distance covered by the object before reaching the ground
(c) The speed of the object just before reaching the ground
Data
 126 
 cos 0 = 35m / s ; U y =35sin0º = 0; h = S y = 50m
 3 .6 
Ux= 
Solutions
1 2
gt
2
(a) S y = U y t +
50 = 0 +
1
 9.81 t 2
2
 50 


 4.905 
t=
t = 2.078s
(b) Range = Ucosθt
= (35cos0)  2.078
Range = 72.738m
(c) V x = U x = 35sin0=35m/s;
Uy= 0
V2
y
= U2
y
+ 2gS y
= 0 + (2  9.81 50)
=
981
V y = 31.32m/s
V=
(V
2
y
+V 2x
)
V=
(352 + 31.322 )
V=
2205.942
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Engineering science for Automotive & HER Programs Level I
V= 46.97m/s
4. A body is projected from a height of 40m at an angle of 25° to and below the horizontal
with an initial velocity of 20m/s. For the motion of this projectile, determine;
(a) time taken for the body to reach the ground
(b) the horizontal distance covered by the body before reaching the ground
Data
U y = 20sin25 = 8.452m/s; S y = 40m; a y = g = 9.81m/s 2
Solutions
(a) S y = Usinθt +
1 2
gt
2
But Usinθ = U y , thus
40 = 8.45t +
1
 9.81t 2
2
40 = 8.45t + 4.905t 2
Rearranging the expression in the format of a quadratic equation,
4.905t 2 + 8.45t – 40 = 0, then using standard formula
− b  b 2 − 4ac
t=
2a
− 8.45  8.452 − (4  4.905  −40)
t=
2  4.905
=
− 8.45  71.436 + 784.8
9.81
t=
− 8.45  856.356
9.81
t = 2.12S or t = -3.84S
 t = 2.12sec
(b) Range = Ucosθt
Range = (20cos25)  2.12
Range = 38.43m
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Engineering science for Automotive & HER Programs Level I
Exercise
1. A body is projected from the earth’s surface at an angle 50° to and above the horizontal
with an initial speed of 30m/s. For the motion of this projectile, calculate;
(a) the total time of flight
(b) the range
2. A projectile is launched from the top of a 70m high building at an elevation of 20º to
and above the horizontal with a velocity of 25m/s. Calculate,
(a) The maximum height reached above the ground
(b) The total time of flight
(c) The horizontal distance covered
(d) The speed of the projectile just before reaching the ground
3. A motor bike is set to jump over obstacles laid across a distance of 20m. To achieve
this, the starting point of the jump is made into a track sloping at elevated at 40º to and
above the horizontal. In order for this motor bike to successfully jump over the obstacles
in 5seconds, calculate;
(a) The duration of the jump
(b) The speed of the motor bike at the start of the jump
4. A helicopter flying horizontally at a height of 200m above the ground with a speed of
120km/h drops a parcel. Calculate;
(a) The time take for the parcel to reach the ground
(b) The horizontal distance covered by the parcel
5. A man standing on a 55m high overhanging rock throws an object at a speed of 10m/s
second at an angle of 10º to and below the horizontal. For the motion of this parcel,
calculate;
(a) The total time of flight
(b) The horizontal distance covered by the parcel
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Engineering science for Automotive & HER Programs Level I
MOMENTUM
Momentum is defined as quantity of motion possessed by a moving body. It is thus given by the
product of the mass and velocity. Thus,
Momentum = mass × velocity (kg.m/s)
Momentum = mv (kg.m/s)
Conservation of momentum
The principle of conservation of momentum states that the total momentum before impact is
equal to the total momentum after impact or collision. Therefore;
∑ momentum before impact = ∑ Momentum after impact
m1 u1 + m2 u2 = m1 v1 + m2 v2
Where, m1=mass of the first body,
m2 = mass of the second body,
u1= velocity of first body before impact
v1, v2=velocity of first and second body after impact, respectively.
Principle of conservation of energy
The principle of conservation of energy states that energy can neither be destroyed nor created,
but can be converted from one form to another. Therefore, during an impact, the initial kinetic
energy;
E1>E2, But this is not in agreement with the principle of conservation of energy. The corrected
expression is therefore as follows:
KE1=KE2 + sound Energy + heat energy
Types of collisions
When two bodies with different momentums collide, there are two possible outcomes that might
be expected. First, the bodies might stick together and move away as a unit. Second, the bodies
might separate and move away with different velocities. Therefore, the three types of collisions
are as follows:
(i)
Inelastic collisions
These are collisions in which the two bodies stick together and move away with a
common velocity afterwards. In such collisions, the initial kinetic energy is not equal
to the final kinetic energy, and only the momentum is conserved. In order to solve
problems involving such collisions, only the principle of conservation of momentum is
needed since kinetic energy is not conserved. Thus;
m1 u1 + m2 u2 = m1 v1 + m2 v2
But since v1 = v2 ,
m1 u1 + m2 u2 = (m1 + m2 )V
Where, V = common velocity or velocity with which the two bodies move away
together as a unit in m/s.
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Engineering science for Automotive & HER Programs Level I
(ii)
Elastic collisions
An elastic collision is one in which the initial kinetic energy before impact is equal to
the final kinetic energy after collision, and the momentum is conserved. Therefore,
when dealing with problems on elastic collisions, the following expressions are used:
m1 u1 + m2 u2 = m1 v1 + m2 v2
And the conservation of kinetic energy expression,
e=
v1 − v2
−(u1 − u2 )
𝑜𝑟 e =
v1 − v2
u2 − u1 )
Where e = the coefficient of restitution. For elastic collisions, e = 1
(iii)
Partially elastic collisions
These are collisions which are neither elastic nor inelastic. The equation of
conservation of momentum and that of conserving kinetic energy are required in
dealing with problems. The coefficient of restitution is between 0 and 1.
The coefficient of restitution for each of the collisions thus stands as follows;
Elastic collisions: e = 1
Inelastic collsions: e = 0
Partially elastic collisions: 0>e<1
Examples
1. A body of mass 0.5kg initially running at a velocity of 10m/s due east collides with another
body of mass 1.2kg moving at 7m/s in the opposite direction. Assuming that the two
bodies stick together after the collision and move away with a common velocity, determine
the magnitude of the velocity with which the two bodies move.
Data:
m1=0.5kg; u1=10m/s; m2=1.2kg; u2= −7m/s; v=?
Solution
m1 u1 + m2 u2 = (m1 + m2 )v
(0.5 × 10) + (1.2 × −7) = (0.5 + 1.2)v
5 − 8.4 = 1.7v
𝐯 = −𝟐𝐦/𝐬 or 2m/s due west
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Engineering science for Automotive & HER Programs Level I
2. An object initially moving at 30m/s has a mass of 4kg. It collides with another body of
mass 10kg moving in the same direction with a velocity of 4m/s. Determine the resulting
velocities if the collision is;
(a) Elastic
(b) Has a coefficient of restitution of 0.6
Data
U1=30m/s; m1=4kg; u2=4m/s; m2=10kg; e = 1 for elastic collisions
Solutions
(a)
m1 u1 + m2 u2 = m1 v1 + m2 v2
(4 × 30) + (4 × 10) = 4v1 + 10v2
2v1 + 5v2 = 80
Equation (1)
Also employing the expression for the coefficient of restitution
e=
v1 − v2
−(u1 − u2 )
1=
V1 −V2
−(30−4)
V1 – V2 = −26
V1 = V2 − 26
Equation (2)
Substituting equation (2) in equation (1) gives the following:
2 (V2 − 26) + 5V2 = 80
7V2 = 132
V2 = 18.857m/s
Substituting in equation (2) for V2 gives:
V1 = 18.857 − 26
V1 = − 7.143m/s
(b)
m1 u1 + m2 u2 = m1 v1 + m2 v2
(4 × 30) + (4 × 10) = 4v1 + 10v2
2v1 + 5v2 = 80
Equation (3)
Applying the expression for the coefficient of restitution:
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e=
v1 − v2
−(u1 − u2 )
0.6 =
V1 −V2
−(30−4)
V1 − V2 = − 15.6
V1 = V2 – 15.6
Equation (4)
Substituting equation (4) in equation (3):
2 (V2 – 15.6) + 5V2 = 80
7V2 = 111.2
V2 = 15.886m/s
Substituting for V2 in equation (4):
V1 = 15.886 – 15.6
V1 = 0.286m/s
3. A gun of mass 1.5kg fires a bullet of mass 80g and recoils with a velocity of 3m/s. It
then hits a block of mass 3kg and remains embedded in the block thereby allowing the
block and the bullet to move away together. Determine;
(a) The velocity with which the bullet leaves the gun
(b) The velocity with which the block and bullet move away together
Data;
m1 = 1.5kg; m2 =
80g
1000
= 0.08kg; u1= 0; u2 = 0m/s; v1 = − 3m/s; v2=?
Solutions;
(a)
m1 u1 + m2 u2 = m1 v1 + m2 v2
(1.5 × 0) + (0.08 × 0) = (1.5 × −3) + 0.08v2
0 = 4.5 + 0.08v2
v2 =
4.5
0.08
𝐯𝟐 = 𝟓𝟔. 𝟐𝟓𝐦/𝐬
(b)
Data;
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m2 = 0.08kg; u2 = 56.25m/s; m3 = 3kg; u3 = 0; v=?
Solution
m2 u2 + m3 u3 = (m2 + m3 )v
(0.08 × 56.25) + (3 × 0) = (0.08 + 3)v
4.5
3.08v
=
3.08
3.08
v = 1.461m/s
EXERCISE
1. A body of mass 40kg is moving at a velocity of 10m/s due east. It collides with another
body of mass 25kg moving at 5m/s in the same direction. Assuming that the two bodies
stick together after the collision, determine the common velocity with which the two
bodies move away.
2. An automobile of mass 750kg initially moving at 72km/h rams into a truck of mass
15tonne initially at rest. If the two bodies separate and the coefficient of restitution for
the collision is 0.7, determine the resulting velocities of the two bodies.
3. During a head-on collision, a body of mass 5.5kg initially running at 20m/s due east
collides with another body of mass 12kg initially running at 8m/s due west. Calculate the
resulting velocities if the collision is;
(a)
(b)
Elastic
Inelastic
4. A gun of mass 3.5kg fires a bullet of mass 50g and recoils at a velocity of 4m/s. The
bullet then hits a stationary plastic can and embeds itself in the can of mass 500g, and
before moving for a while. Compute the;
(a)
(b)
(c)
(d)
Velocity with which the bullet exits the gun
The velocity with which the bullet and can move away together
The time taken for the can to come to rest
The distance slid by the can before coming to rest
5. An object of mass 8kg initially running at 6m/s is hit from behind by an object of mass
5kg moving in the same direction at a speed of 15m/s. Assuming the two bodies
separate after collision is elastic, determine the resulting velocities of the two objects.
NEWTON’S LAWS OF MOTION
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Newton’s laws are based on the study of the interactions between forces and bodies. These
laws are arranged as follows:
Newton’s first law of motion:
This law states that a body remains at rest or continues in its its state of uniform velocity in a
straight line unless acted upon by an externally applied resultant force. This law is also sometimes
referred to as the law of inertia. This is mainly because inertia is the tendency of a body to
maintain its state of motion thus opposing any change that might come.
Newton’s second law of motion:
This law states that the rate of change of momentum is directly proportional to the externally
applied resultant force and acts in the direction of that force. In other terms it states that the
acceleration is directly proportional to the force and inversely proportional to the mass of an
object. Thus,
Fα
mv − mu
t
v−u
)
t
F = m(
But since, a =
v−u
t
Therefore, F = ma
Newton’s third law of motion
This law states that, to every action, there is an equal and opposite reaction. In other words,
when body A exerts a force on body B, body B will also exert a force equal in magnitude to body
A but in the opposite direction.
Examples
1. A body initially moving at a speed of 12m/s is acted upon by a force which causes its
speed to be increased to 30m/s in half a minute. If the body has a mass of 10.5kg,
determine;
(a) The change in momentum
(b) The accelerating force applied on the body
Data
m = 10.5kg; u = 12m/s; v = 30m/s; t = 30seconds
Solutions
(a) Change in momentum = Final momentum – Initial momentum
∆momentum = mv – mu
∆momentum = m (v − u)
∆momentum = 10.5 (30 − 12)
∆momentum = 189kgm/s
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
v−u
)
t
(b) F = m (
30−12
)
30
= 10.5 (
F = 6.3N
2. An elevator of mass 200kg can accelerate at a rate of 4m/s either when descending or
ascending. Determine the force experienced in the supporting wire ropes when the
elevator is;
(a)
Stationed
(b)
Ascending
(c)
Descending
Data;
m = 200kg; a = 4m/s; g=9.81m/s2; Let W = weight or W = mg
Solutions
(a)
The force in the supporting member is called the tension, T. Thus,
T
∑ Upwards forces = ∑ Downwards forces
T=W
T = mg
T = 200 × 9.81
T = 1962N
200 kg
W = mg
(b)
Ascending: The resultant force R opposes the motion. Thus;
Ascending
∑ Upwards forces = ∑ Downwards forces
T=R+W
T
T = ma + mg
T = m (a + g)
T = 200 (4 + 9.81)
200kg
R= ma
T = 2762N
W = mg
(c)
Descending: The resultant force acts in same direction with the tension. Thus;
∑ Upwards forces = ∑ Downwards forces
T+R=W
T + ma = mg
T = m (g – a)
T = 200 (9.81 – 4)
T = 1162 N
Prepared by Mr. Mwaba Ernest
Descending
T
200kg
R = ma
W = mg
35
Engineering science for Automotive & HER Programs Level I
3. Water issues out of a fire hydrant at a rate of 4liters/s and is directed towards a closed
blazing door with a velocity of 15m/s. If the door remains closed and the water does not
rebound while the density of water is 1000kg/m3, determine the force exerted on the door
by the water and give a reason for the nature of your answer.
Data
4 liters/s
Discharge or rate of flow, Q = 1000 = 0.004m3/s;
But density =
mass
Volume
∴ mass flow rate = Density × volumetric flow rate,
•
m = 1000 × 0.004 = 4kg/s
m = 4kg;
t = 1sec;
u =15m/s;
v = 0m/s
Solution
v−u
F = m( t )
0−15
)
1
F=4(
F = − 60N
This is a reaction force exerted by the door on the water
Exercise
1. A particle of mass 30kg initially running at a velocity of 40m/s decelerates uniformly until
it comes to rest in 12seconds. Determine the magnitude of the force responsible for
stopping the motion.
2. During some fire-quenching operation, a hose pipe releases water at a rate of 6kg/s which
directed towards a wall with a velocity of 40m/s. Assuming that the water rebounds with
a velocity of 6m/s, determine the force exerted on the wall by the water.
3. A lift supports a load of mass 700kg and normally accelerates at a rate of 3m/s2 when
ascending and 5m/s2 when descending. Determine the magnitude of the force required
to handle the elevator when the elevator is;
(a)
Stationed
(b)
Ascending
(c)
Descending
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
4. For each of the diagrams shown in the figure below, assuming that the pulleys employed
are frictionless and the mass of the rope is ignored, determine; the resulting tension T
and the acceleration of each of the systems when the masses are released.
T
T
T
1.5kg
3.2kg
Prepared by Mr. Mwaba Ernest
T
4kg
7kg
37
Engineering science for Automotive & HER Programs Level I
CENTROIDS
The center of area for a section of negligible thickness is known as the centroid. In other words
this is the position of the center of gravity for thinner sections. However, sections of significant
thickness have what is known as the center of gravity. In this case, the center of gravity is defined
as a point where the weight of a given object or body is said to be concentrated.
There are many approaches that are used in order to determine the location of a center of gravity.
However, the examples below provide some of the most popular methods employed.
Examples
1. Determine the location of the centroid for the figure shown in the diagram below from
point O.
40mm
120mm
50mm
O
100mm
Solution
Net area,
 areas = (50 X 100) + (40 X 70) = 7800mm
2
Total area for closed fig = 100 X 120 = 12 000mm 2
Total moment of area for the closed fig, about the X-axis = 12000  60 = 720 000mm 3
Total area for removed section = 70  60 = 4200mm 2
Moment of area for the removed section, about the Y-axis = 4200  85 = 357 000mm 3
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
Net moment of area,
Y=
 y  smallarea = 720 000- 147 000 = 573 000mm
 y  smallarea =
 Areas
3
573000
= 73.46mm
7800
Moment of area for removed section, along X-axis;
 x  small, area = 4200  70
= 294 000mm
3
Total moment of area for closed fig, along X-axis = 12 000  50 = 600 000mm
Net moment of area, along X-axis
= 600 000 – 294 000 = 306 000mm
X=
 x  small, area =
 areas
3
3
306000
= 39.23mm
7800
Therefore, location of centroid = 39.23mm, 73.46mm
2.
The thin section shown in the figure below has the area concentrated at some point on
the figure relative to point P. Locate the position of this centroid on the figure.
180mm
50mm
120mm
m
200mm
60mm
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
P
80mm
Solution
Total area of closed fig = 200  180 = 36 000mm 2
Removed area1= 150  40 = 6 000mm 2
; Removed area2 = 60  80 = 4800mm 2
Area of portion removed = (6 000 + 4800) = 10 800mm 2
Net area = 36 000 – 10 800 = 25 200mm 2
 Total  area  y = (36000  100) = 3 600 000mm
3
 Re moved  area  y = (6 000  75) + (4800  40) = 642 000mm
3
 Net.area  y = [  Total.area  y ] – [  Re moved.area  y ]
 Net.area  y
Y=
= 3 600 000-642 000 = 2 508 000mm 3
 Net.area  y =
2,958,000
= 117.38mm
25,200
Net.area
 Total.area  x = 36 000  90 = 3 240 000mm
3
 Re moved.area  x = (6 000  100) + (4800  150) = 720 000 + 720 000
 Re moved.area  x = 1 320 000mm
3
 Net.area  x = [  Total.area  x ] - [  Re moved.area  x ]
= 3 240 000 – 1 320 000
 Net.area  x = 1 920 000mm
X=
 Net.area  x
Net.area
3
=
1,920,000
= 76.19mm
25,200
Location of centroid = 76.19mm, 117.38mm
Prepared by Mr. Mwaba Ernest
40
Engineering science for Automotive & HER Programs Level I
NB: It should be noted that the other method of dividing the figure into small areas can equally
be used and the solution to be obtained should be the same as what has just been calculated
above, as the alternative solution for example 2 illustrates below:
Area (mm 2 )
Area 1
150  80 = 12 000
12 000  75 = 900 000
Area  x (mm 3 )
12 000  40 = 480 000
Area 2
50  180 = 9 000
9 000  175 = 1 575 000
9 000  90 = 810 000
Area 3
70  60 = 4 200
4 200  115 = 483 000
4200  150 = 630 000
Total
 Area = 25 200
 Area  y = 2 958 000
 Area  x
Y =
X=
 Area  y =
 Area
Area  y (mm 3 )
= 1 920 000
2,958,000mm 3
=117.38mm
25,200mm 2
 Area  x = 1,920,000 = 76.19mm
25,200
 Area
centroid= (76.19mm,177.38mm
3. Determine the position of the centroid for the uniform thin section shown in the figure
below if the hole has a diameter of 20mm.
Ø18mm hole
20mm
100mm
20mm
40mm 20mm
20mm
10mm
110mm
Solution
Net area = Total area – area of hole – rectangular area removed - area of triangular corner
Prepared by Mr. Mwaba Ernest
41
Engineering science for Automotive & HER Programs Level I
Net area,
   182

A
=
(
100

120
)
−

 4

1

 − (70  60) −   20  20 
2


= 12,000 – 254.469 – 4200 – 200
 A = 7,345.531mm
2
From above calculation,
Total area figure ignoring removed sections, A1 = 12,000mm2
Area circular hole removed from the left hand region, A2 = - 254.469mm2
Area of rectangle removed from right hand region, A3 = - 4200mm2
Area of triangular portion removed from corner, A4 = 200mm2
EXERCISE
1. Locate the position of the centroid for each of the following sections shown below from
the position marked Q.
50mm
(a)
80mm
90mm
30mm
m
20mm
Q
120mm
Prepared by Mr. Mwaba Ernest
42
Engineering science for Automotive & HER Programs Level I
(b)
20mm
120mm
80mm
70mm
15mm
Q
30mm
120mm
2. Determine the location of the centroid for the thin section figures shown in each of the
following diagrams;
(a)
100mm
20mm
15mm
60mm
45mm
(b)
180mm
15mm
50mm
60mm
∅12mm
12mm
200mm
Prepared by Mr. Mwaba Ernest
40mm
20mm
43
Engineering science for Automotive & HER Programs Level I
EQUILIBRIUM
When two or more forces acting at a point are so arranged that they balance each other, then
the forces are said to be in equilibrium. In other terms, a body is said to be in equilibrium if
under the action of two or more forces there is no resultant out of balance force, or moment.
Conditions for equilibrium
The two conditions for equilibrium are as follows:
1. The summation of forces in any given plane must be equal to zero
2. The summation of moments about any fixed point must be equal to zero
Resultant:
This is a single force which can replace a system of forces acting at a point and still produce the
same effect as those forces.
Resultant
F2
Equilibrant:
F1
This is a single force which can produce equilibrium when added to a system of forces acting at
a point. An equilibrant and a resultant can have the same magnitude but they act in opposite
directions to each other.
Equilibrant
F2
F1
Concurrent forces:
These are forces whose lines of action pass through a common point. In other terms these are
forces which pass through the same point.
F2
F1
F3
Prepared by Mr. Mwaba Ernest
44
Engineering science for Automotive & HER Programs Level I
Coplanar forces:
These are forces which act on the same geometric plane. Coplanar forces can also be defined
as forces which lie on the plane.
Collinear forces;
These are forces which act in the same line.
Parallelogram of forces law:
This law states that if two forces acting at a point are represented in magnitude and direction
by the adjacent sides of a parallelogram, then their resultant can be represented in magnitude
and direction by the diagonal of the parallelogram.
F2
R
F1
Triangle of forces law:
This law states that when three forces acting at a point are in equilibrium, they can be
represented in magnitude and direction by the sides of a triangle taken in cyclic order.
F1
F2
F1
F3
F2
F3
Free body diagram
Polygon of forces law:
This law states that when four or more forces acting at a point are in equilibrium then they can
be represented in magnitude and direction by the sides of a polygon taken in cyclic order. This
law is normally employed when applying the graphical method in which a scale can be selected
to represent a suitable magnitude of a force.
Prepared by Mr. Mwaba Ernest
45
Engineering science for Automotive & HER Programs Level I
EXAMPLES
1. The system of concurrent coplanar forces shown in the diagram below is in equilibrium
under the action of the three forces. Determine the magnitudes of the tensions T1 and
T2.
40°
20°
T2
T1
800N
Method 1
The free body diagram appears as shown in the diagram below,
50°
800N
70°
40°
T1
40°
20°
T2
20°
T1
T2
800
=
=
sin 60 sin 70 sin 50
T1
800
800sin 70
=
 T1 =
sin 60 sin 70
sin 60
𝐓𝟏 = 868.05N
800
Sin 60
=
T2
Sin 50
⟹ 𝑇2 =
800 Sin 50
Sin 60
𝐓𝟐 = 707.64N
Method 2
Applying the resolution of forces method,
T1 Sin 40
40°
T1
T2
20°
T1cos 40
T2 cos 20
800N
Prepared by Mr. Mwaba Ernest
T2 Sin 20
46
Engineering science for Automotive & HER Programs Level I
Resolving forces in the x-axis
T1 Cos 40 = T2 Cos 20 ⟹ T1 =
T2 Cos 20°
Cos 40°
Equation (1)
Resolving forces in the y-axis
T1 Sin 40 + T2 Sin 20 = 800.
Equation (2)
Substituting equation (1) in equation (2) for T1 gives;
T2 Cos 20
) Sin 40 + T2 Sin 20
Cos 40
Cos 20 ×Sin 40
T2 ( Cos 40
+ Sin 20) =
(
T2 =
= 800
800
800
1.1305
∴ 𝐓𝟐 = 707.6415N
Substituting for T2 in equation (1);
T1 =
707.6415 ×Cos 20
Cos 40
𝐓𝟏 = 868.05N
2. A load of mass 30kg is hung with the help of two supports as shown in the diagram
below. The tension in the left hand support is 200N and the system is in equilibrium.
Calculate the tension T and the angle θ.
T
θ
200N
θ
30°
𝜶
W=294.3N
T
60° 200N
30°
30kg
Method 1
Applying the cosine rule on the free body diagram;
T 2 = 2002 + 294.32 − (2 × 294.3 × 200Cos 60)
T 2 = 126,612.49 – 58,860
Prepared by Mr. Mwaba Ernest
Free body diagram
47
Engineering science for Automotive & HER Programs Level I
T = √67752.49
T = 260.293N
The sine rule can now be applicable in calculating for the angle, thus;
260.293
Sin 60
=
Sin 𝛼 =
200
sin 𝛼
200 sin 60
260.293
= 0.6654
𝛼 = 41.715°
θ + 𝛼 = 90°
∴ θ = 90 – 𝛼
= 90 – 41.715
∴ θ = 48.3°
Method 2
T
Tsin θ
θ
Tcos θ
Resolving forces in the y-axis;
200N
30˚
200cos 30
294.3N
Tsin θ + 200sin 30 = 294.3
Tsin θ = 194.3
Equation (1)
Resolving forces in the x-axis;
Tcos θ = 200cos 30
Equation (2)
Dividing equation (2) by equation (1) gives the following;
Tsin θ
194.3
=
T cos θ 200 cos 30
Tan θ = 1.1218
Prepared by Mr. Mwaba Ernest
200sin 30
48
Engineering science for Automotive & HER Programs Level I
θ = tan-1 (1.1218)
θ = 48.285˚
Substituting for θ in equation (1), gives the following;
Tsin 48.285 = 194.3
194.3
T = sin 48.285
T = 260.293N
3. Determine the magnitude and direction of the resultant for the system of forces shown
in the diagram below.
20kN
60kN
70°
50°
80kN
40kN
Solution
Resolving in the x-axis, gives the horizontal component summation,
H = 80cos 0° + 60Cos 50° + 20Cos 110° + 40Cos 270°
H = 111.727N
Vertical components summation, gives
V = 80Sin 0° + 60 Sin 50° + 20 Sin 110° + 40 Sin 270°
V = 24.756N
Resultant
R = √H 2 + V 2
H=111.727N
12.5°
V=24.756N
R = √111.7272 + 24.7562
R = 114.44N
R=114.44N
Direction only requires the angle,
or
tan θ =
θ = tan
V
H
−1
=
24.756
=
111.727
0.2216
0.2216 = 12.5°
and 180 + 12.5 = 192.5°
∴ θ = 12.5° or 192.5°
Prepared by Mr. Mwaba Ernest
R=114.44N
12.5°
H=111.727N
V=24.756N
49
Engineering science for Automotive & HER Programs Level I
4. A jib and tie assembly is employed to lift a load of 20kN as shown in the diagram below.
If the system is in equilibrium, calculate the tensions in the tie and jib.
tie
60°
jib
20kN
40°
Solution
tie
20°
60°
120°
jib
20kN 40°
Applying the sine rule
20
tie
= Sin 40 ⟹ tie =
Sin 20
20 Sin 40
Sin 20
Similarly
jib
20
= Sin 20 ⟹ jib =
sin 120
20 Sin 120
Sin 20
∴ tie = 37.588kN
∴ jib = 50.642kN
EXERCISE
1. Determine the tensions T and F in the system of forces shown in the diagram below if
the system is in equilibrium. (Ans: T = 1141.29N; F =1290.25N)
Prepared by Mr. Mwaba Ernest
50
Engineering science for Automotive & HER Programs Level I
40°
30°
F
T
1400N
2. A lab technician hangs a 2.4kg L.C.D projector with a one of the ropes having a
maximum capacity of 15N and he ties it at an angle of 50° to the vertical as the diagram
illustrates below. Determine the tension T in the other rope and the angle 𝛼.
( Ans: T =18.04N; 𝛼 = 50.4° )
𝛼
50°
15N
T
2.4kg LCD projector
3. Five concurrent coplanar forces act at a point as shown in the figure below. If the
system is no longer in equilibrium, calculate the magnitude and direction of the
resultant.
(Ans: R=164.7N; θ = 30.8° and 210.8°)
50kN
30kN
100kN
20°
50°
80kN
43°
30°
70kN
4. Two forces of magnitudes 60N and 30N are acting at a point, at the same instant. If the
60N load is inclined at 25° to and above the horizontal while the 30N load is inclined at
Prepared by Mr. Mwaba Ernest
51
Engineering science for Automotive & HER Programs Level I
70° to and above the horizontal, calculate the magnitude of the resultant force acting on
this system. (Ans: 83.938N)
5. Two teams battle it out in a tug- of -war such that team A exerts a force of 2.8kN while
team B exerts the same amount of force in the opposite direction. For this simple
system of forces set up, calculate the;
(a) Resultant
(Ans: 0)
(b) Tension in the rope being used (Ans: 5.6kN)
6. A 20kg load is supported between two members such that the angle between the first
member and the horizontal is 60° and the angle between the horizontal and the other
support is 45°. Determine the tensions T, P and W if the system is in equilibrium.
( Ans: 101.56N, P=143.628N and W= 196.2N )
60°
P
T
45°
W
20kg
7. Due to limited Space in the working environment, two loads of weights 30N and 80 N
are supported in equilibrium as illustrated in the figure below. Determine the tensions in
each of the labeled members. (Ans: K= 124.458N; L = 95.34N; M=67.9; J=183.7N)
20°
M
K
L
40°
80N
Prepared by Mr. Mwaba Ernest
30N
30°
J
52
Engineering science for Automotive & HER Programs Level I
FRICTION
Friction is the force which opposes relative motion between two surfaces which are in sliding
contact.
Laws of dry friction
These laws are based on experiments which were conducted between dry and clean surfaces.
They are as follows;
1. Friction always opposes motion (the friction force always acts in the opposite direction
to the direction of the relative motion)
2. The frictional force is directly proportional to the normal reaction between the surfaces
in contact ( F=µN)
3. Friction depends on the type of surfaces in contact (e.g rubber on metal, wood on
plastic etc)
4. Friction depends on the condition or nature of the surfaces involved (e.g smooth, rough,
dirty…)
5. The frictional force is independent of the area of the surfaces in contact
6. At low speeds, the frictional force is independent of the sliding speed
Advantages of friction
(i)
(ii)
(iii)
(iv)
(v)
It helps in the gripping of objects (e.g. Roller on paper in a printer)
It facilitates in the transmission of motion or power
It helps in the braking of machinery
If assists in the generation of useful heat
It enables movement between destinations (e.g. walking, running, rolling etc)
Disadvantages of friction
(i)
(ii)
(iii)
(iv)
(v)
It
It
It
It
It
generates unwanted heat
reduces the efficiency of machinery
causes wear and tear
opposes motion
can sometimes cause noise pollution
EXAMPLES
1. A load having a weight of 200N is pulled on a horizontal floor using a force of 20N
applied horizontally. Determine;
2.
(a) The frictional force
(b) The coefficient of friction between the load and the floor
Prepared by Mr. Mwaba Ernest
53
Engineering science for Automotive & HER Programs Level I
Solutions
W = 200N
20N
F=µN
N
(a) Resolving forces in the X-axis
Ff = 20N
(b) Resolving forces normal to the y-axis
N = 200
But, Ff = µN
∴µ=
=
Ff
N
20
200
𝜇 = 0.1
3. A toolbox having a mass of 10kg is moved along a horizontal surface using an effort
inclined at 30° to and above the horizontal. If the coefficient of friction between the
surface and the toolbox is 0.3, calculate the magnitude of the applied effort when the
toolbox is being;
(a) Pulled
(b) Pushed
Solution
(a) DATA
Weight, W = 10 × 9.81=98.1N; θ=20°; µ= 0.3
98.1
P
P sin 30°
30°
P cos 30°
Ff =µN
N
Resolving forces parallel to the plane
Pcos 30
µN=p cos 30 ⟹ N =
0.3
Resolving forces normal to the plane gives;
N + psin 30° = 98.1
Prepared by Mr. Mwaba Ernest
eq (1)
eqn (2)
54
Engineering science for Automotive & HER Programs Level I
Combining equations (1) and (2) by substituting for N gives,
p cos 30
)
0.3
(
+ P sin 30 = 98.1
98.1
⟹ P = 2.887+sin 30 = 28.966N
(b)
pushing
P sin 30°
P
98.1
30°
Pcos 30
Ff = 0.3N
N
Resolving forces parallel to the plane;
0.3N = P cos 30° ⇒ N =
Pcos 30
0.3
eqn (1)
Resolving forces normal to the plane;
Psin 30° + 98.1 = N
eqn (2)
Substituting eqn (1) in eqn (2) for N and rearranging gives;
cos 30
0.3
P(
∴P=
− sin 30) = 98.1
98.1
2.887− sin 30
P = 41.1N
4. An object of weight 800N is pulled up a plane inclined at 45° to and above the
horizontal using an effort of 750N. Determine the coefficient of friction between the
object and the plane.
N
Ff =µN
W=800N
45°
750N
45°
800 cos 45°
800 sin 45°
Prepared by Mr. Mwaba Ernest
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Engineering science for Automotive & HER Programs Level I
Resolving forces normal to the plane;
N = 800cos 45°
equation (1)
Resolving forces parallel to the plane,
µN + 800sin 45 = 750
equation (2)
Substituting equation (1) in equation (2) for N
µ(800cos 45) + 800 sin 45 = 750
µ=
750−800 sin 45
800 cos 45
⟹ µ= 0.3
EXERCISE
1. A block of weight 50N is resting on a horizontal surface and is required to be pushed
from rest. If the coefficient of static friction between the surface and the block is 0.1,
calculate the magnitude of the horizontal force required to push the block. (Ans:5N)
2. A laptop having a mass of 1.6kg can be moved on top of a table by applying a push of
6.279N. If the push is parallel to the surface of the table, calculate the coefficient of
friction between the table and the base of the laptop. (Ans:0.4)
3. During an experiment to determine the coefficient of friction between two materials, a
load of weight 600N is positioned on a horizontal surface, and a pulling force of 80N is
applied at 30° to and above the horizontal surface. Calculate the coefficient of friction
between the load and the surface. (Ans:0.12)
4. A component having a mass of 20kg is required to be moved across a horizontal floor.
The coefficient of friction between the component and the floor is 0.2. Calculate the
force inclined at 20° to and above the horizontal required to,
(a) Pull the component (Ans:38.92N)
(b) Push the component (Ans:46.73N)
(c) Pull the component if the force is horizontal (Ans:39.24N)
5.
A tool box of weight 1100N containing computer accessories is positioned on an incline
of 25° to and above the horizontal. If the coefficient of friction between the box and the
floor is 0.3, calculate
(a) The normal reaction between the incline and the box
(Ans:996.938N)
(b)
The effort parallel to plane required pull the box up the incline (Ans:763.96N)
6. A 50N weight wooden block is resting on an incline of 10° to and above the horizontal.
If the coefficient of friction between this block and the plane is 0.3, calculate the force
required to drag the block down the gradient if this effort is applied parallel to the plane.
(Ans: 12.179N)
7. An effort of 20N is required to pull a load of 400N on a horizontal surface. Determine;
(a) The coefficient of friction between the surfaces in contact
(Ans:0.05)
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(b)
The effort inclined at 40° to and above the horizontal required to pull the load
(Ans:25.06N)
8. The system shown in the diagram below is just at the point of allowing the 100N load to
slide down the gradient. The frictionless pulley supports a string which connects the
100N load and the 200N load. If the 100N load is resting on a frictionless plane,
calculate;
(a) The pull exerted on the 200N load
(b) The coefficient of friction between the 200N load and the plane on which it is
supported (Ans:0.25)
200N
Frictionless pulley
100N
30°
9. A ball of weight 240N is supported as shown in the figure below with the help of a
671.30N load at the point of sliding forward. Determine the tensions T1 , T2 , T3 , and
the coefficient of static friction between the 700N load and the surface on which it is
resting. (Ans: T1=240N, T2=313.3N, T3=201.38N, µ=0.3)
50°
673.1N
T2
T3
T1
240N
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WORK, ENERRGY AND POWER
Work;
This is a product of a force and the displacement covered in the direction of the applied force.
Work done = F × 𝒅
Power,
Power is the rate of doing work. It can also be defined as the work done per unit time.
𝐩𝐨𝐰𝐞𝐫 =
𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞
𝐭𝐢𝐦𝐞
But since, work done = F x d substituting in the equation for power gives;
Energy,
Power = Force x velocity i.e power = FV
Energy is the capacity for doing work. In other terms it can be defined as a quantity that has
the capacity to do work. There are many forms of energy which include the following;
(i)
Kinetic energy
Kinetic energy is defined as the energy possessed by a body by virtue of its motion.
It is also defined as the energy possessed by a body in motion. It is given by,
𝟏
𝐦𝐯 𝟐
𝟐
Where, m = mass in kg; v = velocity in m/s
𝐊. 𝐄 =
(ii)
Gravitational potential energy
This is the energy possessed by a body due to its height above the datum. It is also
referred to as potential energy. There are many form of potential energy, which
include chemical, electric, atomic etc.
The gravitational potential energy is determined by
P.E = mgh
Where, m = mass in kg; g = acceleration due to gravity in m/s2; h = height above
the datum
(iii)
Rest energy
This is the amount of energy equivalent to the rest mass of a body. In simpler terms
this is the energy possessed by a body due to its mass alone. It is determined by,
Einstein’s formula,
Eo =mo C2
Where, Eo=rest energy; mo = mass of body at rest; C= velocity of light (3 × 108 m/s)
EXAMPLES
1. A force having a magnitude of 70kN is applied on a body through a displacement of
100m within a period of 40seconds. Determine the;
(a) Work done
(b) Power developed
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DATA
F = 70, 000N; d = 100m; t = 50sec
(a) Work done = F x d
= 70, 000 x 100
Work done = 7,000,000J or 7MJ
(b)
Power =
=
work done
time
7,000,000
50
Power = 140, 000Watts or 140kW
2. A load is pulled on a horizontal surface using an effort of 40N applied at 30° to and
above the horizontal through a distance of 120m. Determine the work done in the 120m
long displacement.
40N
30°
40cos30
DATA
F = 40N; θ = 30°; d = 120m
Solution
Work done = Fcosθ x d
= 40cos 30 × 120
Work done = 4156.92 J
3. A body of mass 100kg is supported at height of 60m above the ground. If the body
allowed to fall freely, calculate the;
(a)
(b)
(c)
(d)
Velocity after falling through 40m
kinetic energy after falling through 40m
Striking velocity
Potential energy after falling through 20m
DATA
h = 60m ; m = 100kg;
(a)
Solutions
PE = KE
1
mgh = 2 mv 2
v = √2𝑔ℎ
v = √2 × 9.81 × 40
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v = 28.01m/s
(b)
1
2
K.E = mv 2
1
K.E = 2 × 100 × 28.012
K.E = 39,228J
(c)
KE = PE
v = √2gh
v = √2 × 9.81 × 60
v = 34.31m/s
(c) PE = mgh = 100 × 9.81 × ( 60-20)⟹ PE = 39240J
4. A mass is acted upon by a force which produces a work of magnitude 90MJ and the
power output is noted to be 600kW. If the force acts on the body through a
displacement of 1.2km, calculate the,
(a) Duration of the process
(b) Magnitude of the applied force
DATA
Work done = 90,000,000J; power = 600,000W; d = 1.2 x 1000 = 1200
Solutions
(a) Power =
work done
power
time =
=
work done
time
90,000,000
600,000
time = 150sec
(b)
Work done = force × displacement
Force =
=
work done
displacement
90,000,000
1200
Force = 75,000N
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EXERCISE
1. A force of 50kN acts on a body through a displacement of 80m in the direction of the
applied force. If the time taken for the displacement to be covered is 1minute, calculate;
(a) The work done
(Ans: 4MJ)
(b) The power developed
(Ans: 66.67kW)
2. When an object is acted upon by a force which produces 7.4MW of power, the time
taken is noticed to be 80seconds. If the object is caused to move at a constant velocity
of 90km/h, calculate the;
(a) Work done
(Ans :592MJ)
(b) Force applied (Ans:296kN)
3. A lab technician drags a box containing monitors and keyboards using an effort of 485N
on a horizontal floor. If the bar being used for pulling the box is at an angle of 40° to
and above the horizontal while the power developed is 22kW, calculate the;
(a) Constant velocity at which the box may be moving (Ans: 59.2m/s)
(b) Work done in 3minutes
(Ans:3.96MJ)
4. A laptop of mass 2.1kg is resting on a table of height 1.2m. If the table is heavily
bumped into and the laptop falls off, determine the,
(a) Velocity with which it lands on the floor
(Ans: 4.85m/s)
(b) Potential energy just before it started falling freely
(Ans: 47.72 J)
5. Two desktops each having a mass of 4.5kg are raised to a height of 6m at a constant
velocity of 10m/s. Calculate the work done and the power developed. (Ans: 529.74J;
882.9W)
6. An object covers a distance of 300m is 10seconds while moving between two
destinations at a constant speed. If the kinetic energy developed by the particle is
1350J, calculate the mass of the object. (Ans: 3kg)
7. A mass of 20kg is supported at a height of 200m above the ground. If it is thrown
vertically downwards with a velocity of 25m/s and the air resistance is ignored,
calculate;
(a) Kinetic energy of the mass when reaching midway (Ans: 25.87kJ)
(b) Kinetic energy just before reaching the base
(Ans: 45.49kJ)
8. A helicopter hovering at a height of 220m height and moving at 60m/s drops an object
of mass 70kg. Assuming that the air resistance is ignored, calculate the;
(a) Potential energy of the object just before it starts falling
(Ans:151,074J)
(b) kinetic energy of the object 20m before the end of the fall (Ans:137,340J)
9. During some testing process on a compact disc player, a disc of radius 6cm is rotated at
500rpm. If the mass of the disc is 2.5grams, calculate the kinetic energy of the
disc.(Ans:0.0123J)
10. Two lab technicians exert a force of 28kN inclined at θ° to the horizontal on to an object
and cause it to cover a distance of 100m in 1.5minutes. If the work output is 990kJ,
calculate;
(a) Power developed (Ans:11kW)
(b) Angle θ°
(Ans:69.3)
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HEAT
Heat is defined as a form of energy which is transmitted between two bodies at different
temperatures. It entails that heat can only be transferred when there is a temperature gradient.
Energy can be gained or lost and so heat can be gained or lost through the various heat
transfer methods.
Temperature
Temperature is the degree of hotness or coldness of a body measured against some standard
hotness. It is therefore important to note that there is a difference between heat and
temperature. There can be no heat transfer when the temperature is constant, but there can be
a temperature difference when a process is carried out under constant heat conditions.
Effects of heat;
There are many effects which heat brings about and some of them include;
(i)
Change of state
(ii)
Change in colour
(iii) Change in dimensions
(iv) Change in electrical properties
(v)
Change in shape
Temperature conversions
In order to measure the temperature of a substance, a suitable scale is selected which can be
understood and interpreted. The three common temperature scales are Kelvin, Celsius and
Fahrenheit. Below are the relationships between the temperature scales;
Celsius and Kelvin;
When converting the temperature from Celsius to Kelvin the following expression is employed,
T = t + 273
Where, T = temperature in Kelvin or absolute temperature
t = temperature in degrees Celsius
When the temperature is to be converted from Kelvin to degree Celsius the expression
becomes,
t = T – 273
Fahrenheit and Celsius;
The temperature can be converted from degrees Fahrenheit to degrees Celsius using the
following formula,
𝟓
t = (𝐅 − 𝟑𝟐) 𝟗
Where, F= temperature in degree Fahrenheit and t= temperature in degree Celsius
When the temperature is to be converted from Celsius to Fahrenheit the formula is,
𝟗
F = 𝟓 t + 32
EXAMPLES
1. Convert each of the following to the stated temperature scales,
(a)
40°C to K
(b) 600k to °C
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(c)
(d)
273°C to K
212°F to °C
Solutions
(a)
T = t + 273
T = 40 + 273 = 313K
(b)
t = T – 273 = 600 – 273 = 327°C
(c)
T = t + 273 = 273 + 273 = 546K
(d)
t = (F − 32) 9 =(212 – 32)
5
5
9
= 100°C
2. Calculate the equivalent temperature in the required temperature scale for each of the
following;
(a)
373K to °F
(b)
– 279.4°F to K
(c)
0K to °C
(d)
37.778°C to °F
Solutions
(a)
t = T – 273 = 373 - 273 = 100°C, thus t = 100°C
9
9
F = 5 t + 32 = 5 × 100 + 32 = 212°F
(b)
t = (F – 32)
5
9
= (- 279.4 – 32)
5
9
= - 173°C
T = t + 273 = -173 + 273 = 100K
(c)
t = T – 273 = 0 – 273 = - 273K
(d)
F = 5 t + 32 =
9
9
5
× 37.778 + 32 = 100°F
EXERCISE
1. Convert each of the following to the stated temperature scales;
(a)
250°C to K (Ans:523K)
(b)
800°C to °F (Ans: 1472°F)
(c)
200°F to °C (Ans: 93.3°C)
(d)
350K to °C (Ans: 77°C)
(e)
1000K to °C (Ans: 727°C)
2. Carry
(a)
(b)
(c)
out the following temperature conversions;
700K to °F (Ans: 800.6°F)
30°F to K
(Ans: 271.89K)
600°F to K (Ans: 588.56K)
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(d)
30K to °F
(Ans: - 405.4°F)
(e)
90K to °F
(Ans: - 297.4°F )
3. Calculate the equivalent temperature in the stated temperature scale for each of the
following,
(a)
1200K to °C (Ans: 927°C)
(b)
3700°F to °C (Ans: 2037.78°C )
(c)
340K to °C
(Ans: 67°C)
(d)
450°C to °F (Ans: 842°F)
(e)
60°F to K
(Ans: 288.56)
(f)
700°C to °F (Ans: 1292°F)
Linear expansion
One of the effects of heat is the change in dimensions of a substance which is seen in the linear
expansion of the materials involved. The coefficient of linear expansion is needed in finding the
change in length experienced by a substance.
Coefficient of linear expansion;
This is the amount by which a material changes its dimension per unit change in temperature.
It is denoted by the Greek letter 𝜶.
The coefficient of superficial expansion deals with the area and is therefore twice the coefficient
of linear expansion, i.e
𝜷 = 2𝜶
The coefficient of cubical expansion deals with the volume and is three times the coefficient of
linear expansion. Thus,
𝜸= 3𝛂
Change in length;
The change in length due to a rise or fall in temperature is given by,
𝚫L = 𝛂𝐋𝐨 (𝐭 𝟐 − 𝐭 𝟏 ) or ΔL = 𝜶𝐋𝐨 𝚫𝐭
Where t2= final temperature, t1 = initial temperature, Lo =original length
The thermal strain is given by,
Strain = 𝛂(𝐭 𝟐 − 𝐭 𝟏 )
EXAMPLES
1. A wire of length 2.5m initially at a temperature of 40°C is heated until the temperature
becomes 330°C. If the coefficient of linear expansion for the wire is 0.000003/°C,
calculate;
(a)
Thermal strain
(b)
Change in length
(c)
Final length
DATA
LO = 2.5m; t1=40°C; t2= 330°C; 𝛼 = 0.000003/°C
Solutions
(a)
Thermal strain = 𝛼(t2 – t1 )
Thermal strain = 0.000003(330 – 40) =0.00087
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(b)
Change in length, ΔL = 𝛼Lo (𝑡2 − 𝑡1 )
ΔL = 0.000003 × 2.5 (330 – 40)
𝚫𝐋 = 0.002175m
(c)
L1 = LO + ΔL = 2.5 + 0.002175 = 2.502175m
2. A steel component of cross sectional area 800mm2 and young’s modulus 450GPa is
heated from 50°C to 620°C. If the coefficient of linear expansion for the steel is
0.00000012/°C and its length was 1.25m initially, calculate;
(a)
Change in length
(b)
Stress in the component
(c)
Force produced in the component if it is constrained
DATA
A=
800mm2
10002
= 0.0008m2; E = 450× 109 ; t1=50°C; t2 = 620°C; 𝛼=0.00000012 /°C
Lo = 2.5m
Solutions
(a)
∆L = αLo (t 2 − t1 ) = 0.00000012 × 1. 25 (620 – 50) = 0.0000855m
(b)
Strain = α(t 2 − t1 ) = 0.00000012 (620 – 50) = 0.0000684
Young’s modulus, E =
stress
Strain
⟹ stress = E × strain
Stress = (450 × 109 ) × 0.0000684 = 30,780,000N/m2
(c)
Force = stress × Area = 30,780,000 × 0.0008 = 24624N
3. When a metal rod is heated until the temperature is raised by 510°C the length
becomes 5.000618m. If the stress induced under these conditions is 20N/mm2,
calculate;
(a)
The young’s modulus
(b)
The original length
DATA
L1 = 5.000618m; 𝛼 = 2.6 × 10−7/°C;
Change in temperature, t2 – t1 = 510°C; Stress = 20N/mm2 = 20 × 106 N/m2
Solution
(a)
Strain = 𝛼 (t 2 − t1 ) = 2.6× 10−7(510) = 0.0001236
E=
(b)
stress
strain
Strain =
=
20×106
=
0.0001236
1.5× 𝟏𝟎𝟏𝟏 Pa
L1 − Lo
Lo
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Making original length the subject of the expression gives,
L
5.000618
∴ Lo = (strain1 + 1) = (0.0001236+1) = 5m
EXERCISE
1. A metallic wire of length 4.35m and coefficient of linear expansion 5.1× 10−6/°C is
subjected to an amount of heat which causes its temperature to change from 25°C to
910°C. Calculate;
(a)
Thermal strain
(Ans: 0.00451)
(b)
Change in length (Ans: 0.196m)
(c)
Final length
(Ans: 4.3696m)
2. A rod of length 6m is heated from 27°C to a final temperature of 830°C. If the length of
the rod is increased by 0.05mm during this process, calculate the coefficient of linear
expansion for the material of the rod. ( Ans:1× 10−8/°C)
3. An alloy of diameter 50mm is heated from some unknown temperature until the final
temperature becomes 530°C. The coefficient of linear expansion for the alloy is
1.3 × 10−7 /°C while the diameter of the alloy is 50mm and its original length 3m. If the
heating process causes an expansion of 0.002mm, calculate;
(a)
The initial temperature of the alloy (Ans:524.872°C)
(b)
The strain experienced by the alloy (Ans:6.67× 10−7)
(c)
The force produced in the alloy if its modulus of elasticity is 200GPa
(Ans:261.8N)
4. A 3m long beam is held between two rigid supports while its temperature is raised from
32°C until the temperature is reached when the length becomes 3.00007m. If the
coefficient of linear expansion for the material of the beam is 5× 10−7 /°C and the
young’s modulus for the material is 400GPa while the area is 600mm2, calculate the;
(a)
Final temperature
(Ans:78.67°C)
(b)
Thermal strain
(Ans:0.0000233)
(c)
Force required to overcome the Resultant produced (Ans:5599.2N)
5. The temperatures at the beginning and end of a heating experiment conducted on some
specimen are 50°C and 480°C. If the coefficient of linear expansion for the specimen is
2.6× 10−8/°C, and the length is increase by 0.00001677m, calculate the original length
of the specimen. (Ans: 1.5m)
6. A wire employed in some sensitive work operates in an environment which is maintained
between temperatures 17°C and 60°C while the maximum allowable expansion for this
material is 0.001mm. If the provision for installing this wire only allows for a length of
80cm wire to be fitted, determine the coefficient of linear expansion required for the
material of the wire to satisfy the above conditions. ( Ans:2.9× 10−8/°C)
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Sensible heat and latent heat
This is the amount of heat required to change the temperature of a substance. It is given by;
QS = mc (t2 – t1)
Where, m =mass in kg; c = specific heat capacity; t2= final temperature; t1=initial temperature
QS =sensible heat in Joules (J)
Specific heat capacity;
This is the amount of heat required to change the temperature of a unit mass of a substance by
one degree without a change of state. The unit for specific heat capacity is J/kgK.
Latent heat
The latent heat is the amount of heat required to change the state of a substance without a
change in temperature. The latent heat comes in two categories, i.e;
(i)
Latent heat of fusion;
The latent heat of fusion is the quantity of heat required to change the state of a
substance from solid to liquid or liquid to solid. It is determined by,
QL =m × Lf
Where, m = mass in kg
Lf = Specific latent heat of fusion
QL= latent heat of fusion in Joules (J)
Specific latent of fusion;
This is defined as the amount of heat required to change the state of a unit mass of
a substance from solid to liquid or liquid to solid. It is denoted by Lf as the preceding
expression indicates.
(ii)
Latent heat of vaporization;
The latent heat of fusion is the amount of heat required to change the state of a
substance from liquid to gas or gas to liquid without a change in temperature. It is
determined by;
Qv = m × LV
Where, m = mass in kg
QV = Latent heat of vaporisation
LV = Specific latent heat of vaporization
Specific latent heat of vaporization;
This is the amount of heat required to change the state of a unit mass of a
substance from liquid to gas or gas to liquid. It is denoted by LV and the unit is J/kg.
EXAMPLES
1. Calculate the amount of heat required to raise the temperature of 3kg of some
metal from 20°C to 330°C if the specific heat capacity of the metal is 420J/kgK.
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DATA
m = 3kg; t1= 20°C; t2=330°C; c = 420J/kgK
Solution
Q = mc (t2 – t1) = 3 × 420(330 – 20) = 390,600J
2. A cable of mass 1.8kg experiences an overload in the amount of voltage across
it, which eventually causes it to heat up and the temperature is raised from 18°C
to 110°C. If the amount of heat generated by this cable is 120kJ, calculate the
specific heat capacity of the cable’s material.
DATA
t1=18°C; t2 = 110°C; Q = 120,000J; m=1.8kg; c = ?
Solution
Q = mc (t2 – t1)
120,000 = 1.8 × c (110 – 18)
120,000= (1.8 × 92) c
120,000
C = 165.6
C = 724.64J/kgK
3. An aluminum plate of mass 2.4kg at a temperature of 425°C is plunged in oil of
mass 3.6kg which is at a temperature of 60°C. The specific heat capacity for
aluminium is 920J/kgK and the specific heat capacity for the oil is 3200J/kgK.
If the losses to the environment are ignored, calculate the final temperature of
the mixture.
DATA
Aluminium;
mal = 2.4kg; t al = 425°C; Cal=920J/kgK;
Oil;
moil = 3.6kg; t oil = 60°C; Coil = 3200J/kgK
Solution
Heat lost by aluminium;
Qlost = mal Cal (t al – t) = 2.4× 920 (425 − t)
Qlost = 938,400 – 2208t
Heat gained by oil;
Q gain = moil Coil (t − t oil ) = 3.6 × 3200 (t – 60)
Q gain = 11,520t – 691,200
Heat lost by aluminium = Heat gained by oil
Qlost = Q gain
Equating the two expressions
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938,400 – 2208t = 11,520t – 691,200
2208t + 11,520t = 938,400 + 691,200
13,728t = 1,629,600
t=
1,629,600
13,728
t = 118.7°C
4. A fuse for some plug of a computer can handle melt at 125°C. The specific heat
capacity of the fuse is 800J/kgK and the specific latent heat of fusion for the fuse
is 100KJ/kg. If the fuse is at a temperature of 33°C, and it has a mass of
10grams, calculate the amount of heat energy required for the fuse to melt
completely.
DATA
10
t1 = 33°C; t2 = 125°C; C = 800J/kgK; m = 1000 = 0.01kg; Lf = 100,000J/kg
Solution
Sensible heat,
QS = mc (t2 – t1) = 0.01× 800 (125 – 33) ⟹ QS = 736J
Latent heat of fusion;
QL= mLf = 0.01 × 100,000 = 1000J
Total heat energy required;
Q = QS + QL = 736 + 1000
Q =1736J
EXERCISE
1. Calculate the amount of heat required to raise the temperature of a substance whose
specific heat capacity is 400J/kgK and mass 7kg from 110°C to 890°C. (Ans: 2.184MJ)
2. A metal of mass 20g is heated from 50°C to 550°C. If the amount of heat used in
achieving this temperature rise is 78kJ calculate the specific heat capacity for the metal.
(Ans: 780J/kgK)
3. A steel component of specific heat capacity 320J/kgK is heated from 15°C to 295°C. If
the amount of heat spent in the process is 985.6kJ, calculate the mass of the
component. (Ans:11kg)
4. A fuse of mass 10g initially at 22°C can melt when the temperature rises to 115°C. The
specific heat capacity for the fuse is 700J/kgK and the specific latent heat of fusion is
4.5kJ/kg. Determine the amount of heat required to melt this fuse completely.
(Ans: 696J)
5. A lead plate at of mass 3kg at a temperature of 305°C is plunged in water of mass 4.1kg
which is at a temperature of 25°C. The specific heat capacity for the water is 4.2kJ/kg°C
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and lead 0.13kJ/kg°C. If the losses to the environment are ignored, calculate the
equilibrium temperature reached by the mixture. (Ans: 31.2°C)
6. Determine the amount of heat energy required to melt 1.7kg of some substance which
is at 43°C and whose melting point is 940°C. The specific heat capacity for the
substance is 510J/kgk while the specific latent heat of fusion is 70MJ/kg. (Ans:119.8MJ)
7. Ice of mass 4kg at – 6°C is required to be converted into water at 95°C. If the specific
latent heat of fusion for ice is 335kJ/kg and the specific heat capacity for ice is 2.1kJ/kg
while the specific heat capacity for water is 4.2kJ/kg. (Ans:2986.4kJ)
8. In order for some system to operate at a given temperature of 30°C of mass 2kg at 5°C
is mixed with another body of water at 80°C. If the specific heat capacity for water is
4200J/kgK and the losses to the environment are ignored, calculate the mass of the
other body of water at 80°C which should be added in order for the final temperature of
30°C to be attained. (Ans: 1kg)
GAS LAWS
The gas laws are based on the behaviour of gases once they are subjected to different heat
conditions. It should be noted that these laws apply on an ideal gas and the parameters are
supposed to be absolute in each case. These laws are as follows;
Boyle’s law;
This law states that, for a fixed mass of an ideal gas at a constant temperature, the absolute
pressure is inversely proportional to the volume. It is applicable for a process which is carried
out at a constant temperature which is referred to as an isothermal process. Thus,
1
P α V or P1V1 = P2V2
Charles’ law;
This law states that, for a fixed mass of gas at a constant pressure, the volume is directly
proportional to the absolute temperature. This law is applicable for an ideal gas which is
conducted at a constant pressure and such a process is called an isobaric process.
Mathematically, it is expressed as;
V
V
V α T or T1 = T2
Ideal gas law;
1
2
The absolute pressure of a fixed mass of gas is directly proportional to the absolute
temperature and inversely proportional to the volume. It is thus a combination of Charles’s law
and Boyle’s law.
P1 V1 P2 V2
=
T1
T2
Pressure law;
The absolute pressure is directly proportional to the absolute temperature of a given mass of
gas at a constant volume. A process which is carried out at a constant volume is called an
isochoric process and this marks the condition under which the pressure law is applicable. It is
given by,
P1 P2
=
T1 T2
Prepared by Mr. Mwaba Ernest
70
Engineering science for Automotive & HER Programs Level I
Gas constant, R;
The mass of a gas is constant and by multiplying it with another constant it gives the
expression for the ideal gas law as follows;
PV
= mR
T
Where, R = gas constant in J/kgK; m = mass in kg
EXAMPLES
1. A gas occupies a volume of 0.004m3 at 25°C, if the pressure is kept constant, calculate
the volume occupied by the gas when the temperature is 260°C.
DATA
V1=0.004m3; T1= 25°C + 273 = 298K; T2 = 260°C + 273 = 533K
Solution
V1
V
= 2 ⟹ V2 =
T1
T2
V1 T2
T1
=
0.004×533
298
∴ V2=0.00715m3
2. A quantity of gas initially at an absolute pressure of 180kPa is heated in such a manner
that the temperature remains constant throughout the process. If the gas expands
according to the ratio 1:10, calculate the final absolute pressure of the gas.
DATA
V
1
P1= 180,000Pa; V1 = 10 ∴ V1 = 1; V2 = 10
2
Solution
P 1 V1 = P 2 V2
P V
180,000×1
P2 = V1 1 =
= 18000Pa
10
2
3. A gas of volume 0.1m3 at a pressure of 5.2bar and temperature 600°C is heated until
the volume becomes 1.8m3 and the pressure 0.5bar. Calculate temperature at the end
of the heating process.
DATA
V1 = 0.1m3; P1 = 5.2bar = 5.2× 105 Pa; T1= 600 + 273 = 873K; V2 = 1.8m3;
P2 = 0.5bar = 0.5× 105 Pa
Solution
P1 V1
T1
=
P2 V2
T2
⟹ T2 =
P2 V2 T1
P1 V1
=
(0.5 ×105 )×1.8×873
=1510.96K
(5.2×105 )×0.1
∴ t2= T2 – 273 = 1510.961 – 273 = 1237.96°C
4. An unknown mass of air is contained in a vessel of volume 2.1m3 at an absolute
pressure of 120kPa and temperature 74°C. If the constant for air is 0.287kJ/kgK,
calculate the mass of air contained in the vessel.
DATA
V=2.1m3; P=120kPa = 120,000Pa; T=74+273 = 347K; R=0.287kJ/kgK=287J/kgK
Prepared by Mr. Mwaba Ernest
71
Engineering science for Automotive & HER Programs Level I
Solution
PV
= mR
T
m=
PV
RT
=
120,000×2.1
287×347
=2.53kg
EXERCISE
1. Oxygen is contained in a rigid cylinder at a temperature of 50°C and absolute pressure
1.3MPa. If the temperature is raised to 190°C, calculate the final absolute pressure of
oxygen. (Ans:1.86MPa)
2. A quantity of gas which is left exposed to the rays of the sun for some time is noticed to
have increased its temperature from 13°C to 58°C. If the expansion obeys Charles’ law
and the volume, when the temperature has risen to 58°C, is 2.5× 10−3 m3 , calculate the
initial volume of the gas. (Ans:2.16 × 10−3m3)
3. During the compression process carried out on 1m3 of air at 60kPa and temperature
110°C, the temperature is increased to 734°C. If the volume at the end of the
compression is 0.02m3 calculate the final pressure of the gas. (Ans:7.89MPa)
4.
An adjustable cylinder is used to support a gas which expands according to the law
PV=constant. If the pressure and volume at the beginning of the process are 100kPa
and 1.82× 10−4 m3, calculate the volume of the gas when the pressure has been
reduced to 15kPa. (Ans:1.21 × 10−3m3)
5. An experiment was carried out on 0.5kg mass and 0.06m3 volume of some unknown gas
in order to determine its characteristic gas constant. The gas was under a pressure of
3bar and temperature 260°C. Calculate the characteristic gas constant for this quantity
of gas. (Ans: 67.54J/kgK)
6. A compression process is conducted on some air and it takes place according to the
ratio 12:1. The pressure at the commencement of the compression process is 4500Kpa
and temperature 940°C. If the pressure at the end of the compression becomes 70MPa,
calculate the accompanying temperature. (Ans:1299.4°C)
7. A compressor takes in air at 20°C and volume 0.85m3. The air is compressed at a
constant pressure of 1.8bar until the temperature becomes 527°C. Calculate the final
volume of the air at the end of the compression. (Ans: 2.32m3 )
8. Air at a temperature of 50°C, volume 0.006m3 and absolute pressure 7.2bar under goes
some process which causes its temperature to become 760°C and volume 0.004m3.
Calculate the final pressure of the air. (Ans: 3.45MN/m2)
Prepared by Mr. Mwaba Ernest
72
Engineering science for Automotive & HER Programs Level I
ILLUMINATION
Lighting or illumination is generally defined as the deliberate use of light to achieve a practical or
aesthetic effect. Lighting includes the use of both artificial light sources like lamps and light
fixtures, as well as natural illumination by capturing daylight. Light waves travel at a speed of 3
× 108 m/s in a vacuum and this speed varied depending on the media involved. In the study of
illumination, the following definitions and parameters are of great importance:
Solid angle, ω: This is a three dimensional radian. It is also defined as an angle subtended at
the center of a sphere by a part of its surface whose area is equal to the square of the radius.
The SI unit for solid angle is steradian, Sr.
Candela, (Cd): The candela is the luminous intensity, in a given direction, of a source that emits
monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that
direction of 1⁄683 watt per steradian. It is also defined as one sixth of a radiating cavity at a
temperature of freezing platinum. The relationship between candela and luminous flux is as
follows, 1 cd = 4 π lumens
Luminous flux, Ф; This is the light energy radiated out per second from a source in the form of
luminous light waves. In it given by the light energy contained in each unit solid angle of a source
of 1cd. The SI unit is lumen (lm)
Luminous intensity, I: This is the luminous flux radiated out per unit solid angle in a given
direction. It is the solid angular flux density of a source in a particular direction. The SI unit for
luminous intensity if the candela, cd or lumen per steradian (lm/Sr)
Luminous Intensity, I =
Luminous flux
Solid angle
Luminous Intensity, I =
𝚽
𝛚
(cd)
But since the total solid angle for a sphere or in all direction is, ω = 4 π
Therefore, the total luminous flux is, 𝚽 = 4 𝛑 𝐈 (lm)
Mean spherical candle power (M.S.C.P): This is the value of the candle power in all
directions. It is given by,
MSCP =
Total luminous flux
4π
Illumination, E: This is defined as the normal luminous flux per unit area of the illuminated
surface. Thus;
Ilumination, E =
luminous flux
Surface Area
Prepared by Mr. Mwaba Ernest
73
E=
Engineering science for Automotive & HER Programs Level I
Φ
𝐴
[lm/m2 or lux]
Specific output or efficiency: This is the ratio between the luminous flux and the power intake.
It is given by;
Specific output =
luminous flux
power output
Laws of Illumination
There are three main laws of illumination which specify some conditions that are applied on
illumination in relation to some necessary factors. These laws are arranged as follows:
First law of illumination: This law states that the Illumination, E, is directly proportional to the
luminous intensity, I, of a given source. Therefore,
E𝛼I
Inverse square law or second law of illumination: The illumination of a surface is inversely
proportional to the square of the distance between the surface and the source. Therefore;
E𝛼
1
d2
Lambert’s cosine law or third law of illumination: The illumination, E, is directly
proportional to the cosine of the angle between the normal and incident flux. Thus;
E 𝛼 cos θ
Final expression for illumination:
The final expression deduced from all the above laws is therefore as follows:
E=
I cos θ
d2
This is the expression which can be used in many different cases to solve problems on
illumination. Where,
θ = angle between the normal and the incident flux
I = luminous intensity in candela
d = The distance between the illuminated surface and the source of light.
Prepared by Mr. Mwaba Ernest
74
Engineering science for Automotive & HER Programs Level I
Examples
1. A light source emits light waves of luminous intensity 180cd and is located at a point which
is 3m above the ground. Determine the luminance, in lux, of a point;
(a) Directly below the surface
(b) 4m away from the foot of the source
Data
I = 180cd; d = 3m θ = 0˚
Solutions
(a) E =
E=
I cos θ
d2
180×cos 0
32
E = 20lux
Source
(b)
d =4 +3
2
2
2
θ
d = √(42 + 32 )
d
d = √25 = 5
Cos θ =
3
5
3m
= 0.6
4m
E=
I cos θ
d2
E=
180 ×0.6
52
Surface
E = 4.32lux
2. Determine the illumination of a small surface at a distance of 1.5m from an isotropic point
source of luminous intensity 70cd, if;
(a) The surface is normal to the luminous flux
(b) The normal to the luminous flux makes an angle of 30˚ with the light rays
Data
d = 1.5m; I = 70cd; θ = 0˚
Prepared by Mr. Mwaba Ernest
75
Engineering science for Automotive & HER Programs Level I
Solutions
(a) E =
E=
I cos θ
d2
70 cos 0
1.52
E = 31.111 lux
(b) θ = 30˚
E=
I cos θ
d2
E=
70 cos 30
1.52
E = 26.943 lux
3. A lamp emits light of luminous intensity 16cd towards a screen positioned at 1.15m away.
If the illumination produced by this lamp is the same as what is produced by a bulb
positioned at 1.5m away, determine the luminous intensity of the bulb.
Data;
Ilamp = 16cd; dlamp = 1.15; dlamp = 1.5m; θ = 0˚
Solution
E=
I cos θ
d2
E=
16 ×cos 0
1.152
Elamp = 12.0983 lux
Elamp = Ebulb
12.0983 =
I=
𝐼 cos 0
1.52
12.0983 × 1.52
cos 0
I = 27.221cd.
Prepared by Mr. Mwaba Ernest
76
Engineering science for Automotive & HER Programs Level I
Exercise
1. A light source emits 1300lm of light towards a surface whose dimensions are 2m x 1.2m.
Determine the illumination experienced.
2. An office has a lamp which releases luminous flux of intensity 150cd. The normal or
perpendicular distance between the lamp and the top of the desk is 3m. Determine the
illumination experienced at a point;
(a)
Directly below the lamp
(b)
2m away from the top of the desk
3. Determine the illumination experienced when a source of 90cd is positioned at a
distance of 200cm from the surface, given that;
(a)
The surface is perpendicular to the light source
(b)
The surface is inclined at 40˚ to the normal rays
(c)
The angle between the normal and the light rays is 40˚
4. The illumination of a point which is positioned at 4m from a light source is noticed to be
120lux. Determine the luminous intensity of the source if;
(a)
The surface is normal to the source
(b)
The angle between the source and the surface is 30˚
5. A lamp is emitting 1350lm of luminous flux in all directions is suspended 8m above the
working plane. Determine the illumination at a point on the working plane 5m away
from the point directly below the lamp if 5% of the light is absorbed by emissions in the
environment.
Prepared by Mr. Mwaba Ernest